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Related Rates of rocket

  1. Nov 15, 2007 #1
    1. The problem statement, all variables and given/known data
    A rocket is 1/2 miles in the air going 40mi/h a bystander is standing 1 mile away from where the rocket took off straight up. What is the rate of change of the angle that the bystander makes with the rocket.

    2. Relevant equations

    3. The attempt at a solution

    I set up a triangle and got tan(theta)=y. Then I took the derivative and got dtheta/dt * sec^2(theta) = dy/dt. from here i'm kinda lost it seems like i won't be able to do it now because I have to do this without a calculator so how would I calculate the theta at the time where y=1/2 its not a common angle...?
  2. jcsd
  3. Nov 15, 2007 #2
    You're right that it's not a common angle, but do you really need the angle [itex]\theta[/itex]? You're looking for [itex]\sec(\theta)[/itex] so why don't you use the Pythagorean Theorem, figure out what your hypotenuse is, and then you can find [itex]\sec(\theta)[/itex] without having to know [tex]\theta[/tex]
  4. Nov 15, 2007 #3
    Well, one of your problems is that you do not have enough variables. Try defining the 1/2 miles part x and set up a tan theta= x/1 equation to differentiate.
  5. Nov 15, 2007 #4
    How do you figure there aren't enough variables? With the derived equation one can very easily solve for [tex]\frac{d\theta}{dt}[/tex] which is what we desire. Furthermore, you'll notice that what you suggested, is EXACTLY what physstudent1 did.

  6. Nov 16, 2007 #5
    define the 1/2 mile part to be x. Now set up the equation tan(theta)=x/1. Differentiate and get sec^2(theta)*dtheta/dt=dx/dt. Solve the triangle at the time when x = 1/2, and you find that the hypotenuse is sqrt(5)/2. Solve for sec(theta)^2. Plug in your 40 mph for dx/dt and rearrange the varibles.

    Does that make sense, Kerizhn? I don't know that I did it right, but I get an answer for the problem.
  7. Nov 16, 2007 #6
    Remember to divide your dtheta/dt by 60 to get radians per second change.
  8. Nov 16, 2007 #7
    Younglearner, I'm not sure if you're trying to show us a different way of doing things here, but you emulated precisely what physstudent 1 did in the original post, only with a different variable name.

    Furthermore, you just proved that physstudent1 did indeed have enough variables. I'm not exactly sure what the point of just repeating everything we've already done is...
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